Can someone please show me a walk through for $\displaystyle d/dx (arcsin((2x^2+3)/5)))$

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Hi psupernak Originally Posted by psupernak Can someone please show me a walk through for $\displaystyle d/dx (arcsin((2x^2+3)/5)))$ Is is $\displaystyle [arcsin(x)] ' = \frac{1}{\sqrt{1-x^2}}$ and [(2x^2+3)/5]' = 4/5*x Now use the chain rule $\displaystyle d/dx (arcsin((2x^2+3)/5))) = 4/5*x*\frac{1}{\sqrt{1-((2x^2+3)/5)^2}}$ regards Rapha

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